• The number of nodes is in the range [0, 1000]
• 0 <= Node.val <= 1000
• Columns are output left to right; within a column, rows run top to bottom
• Nodes sharing a row and column are ordered by value ascending

Group the nodes into vertical columns and report every node in each column, ordered top-to-bottom by row — and when two nodes share the same row and column, order them by value.

Give each node two coordinates: a column x (root 0, left child x−1, right child x+1) and a row y (root 0, both children y+1).

                1 (x0, y0)
              /          \
       2 (-1,1)            3 (1,1)
        /    \             /    \
  4(-2,2)  5(0,2)     6(0,2)   7(2,2)

  5 and 6 land on the SAME spot (col 0, row 2), from opposite subtrees
  output:  [[4], [2], [1, 5, 6], [3], [7]]

A level-by-level (BFS) walk visits rows in increasing order, so nodes land in each column already sorted by row — that part is free. The trap is the tie: 5 and 6 share column 0 and row 2 (one went left-then-right, the other right-then-left, both landing back at column 0). A left-to-right BFS visits them in tree order — but the problem says order them by value, which has nothing to do with position. No traversal order can produce that; it must be imposed afterward.

Walk in any order (depth-first is simplest here), recording a triple (x, y, value) per node. Sort the whole list by x, then y, then value — a tuple sort compares the first field first and only consults later ones on ties, exactly the precedence wanted — then cut into groups wherever x changes.

nodes = []

def walk(node, x, y):
    if node is None: return
    nodes.append((x, y, node.val))     # column, row, value
    walk(node.left,  x - 1, y + 1)
    walk(node.right, x + 1, y + 1)

walk(root, 0, 0)
nodes.sort()                            # by x, then y, then value — that priority

out = []
for x, _, val in nodes:
    if not out or x != prev_x:          # a new column starts a new group
        out.append([])
        prev_x = x
    out[-1].append(val)
return out

Coordinates for the tree above, then sorted:

1(0,0)  2(-1,1)  3(1,1)  4(-2,2)  5(0,2)  6(0,2)  7(2,2)

sort by (col, row, value):
  (-2,2,4)
  (-1,1,2)
  (0,0,1) (0,2,5) (0,2,6)   <- 1 first by row; then 5 before 6 by VALUE
  (1,1,3)
  (2,2,7)

cut on column change ->  [[4], [2], [1, 5, 6], [3], [7]]

Had 5 and 6 instead held 9 and 5, that column would read [1, 5, 9] — the values decide, not which subtree the node came from.

The walk is O(N); the sort dominates at O(N log N), with O(N) space for the triples. The judgement to keep: if the required ordering isn't fully determined by structure, stop trying to be clever with the traversal — collect coordinates and sort. A tie-break that lives outside the tree (like value) means an explicit sort isn't a failure to find the elegant walk; it is the answer.